Discrete math- Range and preimage of the floor function

(a) To find ther set $g(S)$ we need to compute $g(x)$ for all values of $1\leq x\leq 6$ i.e.
\[g(S)=\{g(x): 1\leq x\leq 6\},\]
where 
\[g(x)=\lfloor \frac{x-1}{2}\rfloor.\]
is the floor function. It returns the largest integer smaller that $\frac{x-1}{2}$. Note also that $g$ is an increasing function. We compute $g$ at the end points of the interval $1\leq x\leq 6$. 
\[g(1)=\lfloor \frac{1-1}{2}\rfloor=0   \text{and}  g(6)=\lfloor \frac{6-1}{2}\rfloor=2.\]
Hence $g(S)$ is the set of all integers between $0$ and $2$: 
\[g(S)=\{0,1,2\}.\]

(b) We have
\[g^{-1}(T)=\{x: g(x)=2\}=\{x: \lfloor \frac{x-1}{2}\rfloor=2\}.\]
We are looking for all $x$ such that the floor of $\frac{x-1}{2}$ is 2. Hence
\[ 2\leq \frac{x-1}{2} <3    \Rightarrow  4 \leq x-1 <6 \]
\[\Rightarrow 5\leq x <7.\]
Thus 
\[g^{-1}(T)=\{x: g(x)=2\}=\{x\in \mathbb{R}: 5 \leq x <7\}=[5,7).\]